Gauss’s LawGauss’s Law

Alan MurrayAlan Murray

Alan Murray – University of Edinburgh

Revision : Vector Dot Revision : Vector Dot (Scalar) Product(Scalar) Product

a

bθa.b = ab cos θ a

b

a.b = ab cos(90) = 0

a

b

a.b = ab cos(0) = aba

a.a = aa cos(0) = a²

In Cartesian co-ordinates, a.b = (ax,ay,az).(bx.by,bz) = axbx + ayby + azbz

Alan Murray – University of Edinburgh

Revision : Vector x ScalarRevision : Vector x Scalar

a

In Cartesian co-ordinates, for example, 2a = 2(ax,ay,az) = (2ax,2ay,2az)

2a -2a

Alan Murray – University of Edinburgh

Gauss’s Law : Crude AnalogyGauss’s Law : Crude Analogy

Try to “measure” the rain on a rainy dayTry to “measure” the rain on a rainy day• Method 1 : count the raindrops as they fall, Method 1 : count the raindrops as they fall,

and add them upand add them up cf Coulomb’s Lawcf Coulomb’s Law

• Method 2 : Hold up an umbrella (a “surface”) Method 2 : Hold up an umbrella (a “surface”) and see how wet it gets.and see how wet it gets.

cf Gauss’s Lawcf Gauss’s Law Method 1 is a “divide –and-conquer” or “microscopic” Method 1 is a “divide –and-conquer” or “microscopic”

approachapproach Method 2 is a more “gross” or “macroscopic” Method 2 is a more “gross” or “macroscopic”

approachapproach They must give the same answer.They must give the same answer.

Alan Murray – University of Edinburgh

Lines of Electric FieldLines of Electric Field

8C

How many field lines cross out of the circle?

8C ⇒ 8 lines16C ⇒ 16 lines

16C32C 32C ⇒ 32 lines

Alan Murray – University of Edinburgh

Lines of Electric FieldLines of Electric Field

8C

How many field lines cross out of the surface?

8C ⇒ 8 lines16C ⇒ 16 lines

16C32C 32C ⇒ 32 lines

Alan Murray – University of Edinburgh

Gauss’s Law : Cartoon VersionGauss’s Law : Cartoon Version The number of electric field lines The number of electric field lines

leaving a closed surface is equal to leaving a closed surface is equal to the charge enclosed by that surfacethe charge enclosed by that surface

ΣΣ(E-field-lines) (E-field-lines) αα Charge Enclosed Charge Enclosed

N Coulombs ⇒ αN lines

Alan Murray – University of Edinburgh

Lines of Electric FieldLines of Electric Field

8C

How many field lines cross out of the surface?

8C ⇒ 0 lines16C ⇒ 0 lines

16C32C 32C ⇒ 0 lines

i.e. charge enclosed = 0

Alan Murray – University of Edinburgh

Gauss’s Law Proper (Gauss’s Law Proper ())

ΣΣ(E-lines) (E-lines) proportional toproportional to (Charge Enclosed) (Charge Enclosed) ∫∫∫∫DD..dsds = = ∫∫∫ρ∫∫∫ρ((rr)dv)dv

= = ∫∫∫ρ∫∫∫ρ((rr)dxdydz)dxdydz ∫∫∫∫DD..dsds = charge enclosed = charge enclosed DD = = εεEE ε= εε= ε00 = 8.85 x 10 = 8.85 x 10-12-12 in a vacuum in a vacuum

Alan Murray – University of Edinburgh

Digression/RevisionDigression/RevisionArea IntegralsArea Integrals

This area gets wetter!

Alan Murray – University of Edinburgh

Area Integrals – what’s happening?Area Integrals – what’s happening?

dsds

RainfallRainfall

This area gets wetter!

Alan Murray – University of Edinburgh

Area Integrals – what’s happening?Area Integrals – what’s happening?

ds

ds

Rainfall Rainfall

Clearly, as the areas are the same, the angle between thearea and the rainfall matters …

Alan Murray – University of Edinburgh

Area Integrals – what’s happening?Area Integrals – what’s happening?

dsds

Rainfall, R Rainfall, R

Extreme casesat 180° - maximum rainfallat 90°, no rainfall

Alan Murray – University of Edinburgh

Flux of rain (rainfall) through an Flux of rain (rainfall) through an area area dsds

FluxFluxrainrain = = RR..dsds• ||RR||××||dsds||××cos(cos(θθ))• Rds cos(Rds cos(θθ))

FluxFluxrainrain = 0 for = 0 for 90° … cos(cos(θθ) = 0) = 0 FluxFluxrainrain = -Rds for = -Rds for 180° … cos(cos(θθ) = -1) = -1 Generally, FluxGenerally, Fluxrainrain = = Rds cos(Rds cos(θθ))

• -1 < -1 < cos(cos(θθ) < +1) < +1

Alan Murray – University of Edinburgh

Area Integrals : Take-home Area Integrals : Take-home messagemessage

Area is a Area is a vectorvector, perpendicular to the , perpendicular to the surfacesurface

Calculating flux of rain, Calculating flux of rain, EE-field or anything -field or anything else thus involves a scalar or “dot” product else thus involves a scalar or “dot” product aa..bb = ab = abcos(cos(θθ))

This is what appears in a surface integral This is what appears in a surface integral of the form of the form ∫∫∫∫DD..dsds, or , or ∫∫∫∫RR..dsds, which would , which would yield the total rainfall on whatever surface yield the total rainfall on whatever surface is being used for integration (here, the is being used for integration (here, the hills!)hills!)

Alan Murray – University of Edinburgh

ρ l Coulombs/m

L

Gauss’s law - ExampleGauss’s law - ExampleLong straight “rod” of chargeLong straight “rod” of charge

Construct a “Gaussian Surface” that reflects the symmetryof the charge - cylindrical in this case, then evaluate ∫∫∫∫DD..dsds

ds

ds

E, D

ds

E, D

r

Alan Murray – University of Edinburgh

ds

E, D

r

EvaluateEvaluate ∫∫ ∫∫DD..dsds

∫∫∫∫DD..dsds = = ∫∫ ∫∫ DD..dsds curved surfacecurved surface

++∫∫ ∫∫ DD..dsds flat end facesflat end faces

End faces, End faces, DD & & dsds are perpendicular are perpendicular• DD..dsds on end faces = 0 on end faces = 0

• ∫∫ ∫∫ DD..dsds flat end facesflat end faces = 0 = 0

Flat end faces do not contribute!Flat end faces do not contribute!

Alan Murray – University of Edinburgh

EvaluateEvaluate ∫∫ ∫∫DD..dsds

∫∫∫∫DD..dsds = = ∫∫ ∫∫ DD..dsds curved surface onlycurved surface only

ρ l Coulombs/m

L

ds

ds

E, D D & ds parallel,D.ds = |D|×|ds| = Dds

Alan Murray – University of Edinburgh

EvaluateEvaluate ∫∫ ∫∫DD..dsds

∫∫ ∫∫ DD..dsds curved surface onlycurved surface only = = ∫∫ ∫∫ Dds Dds

ρ l Coulombs/m

L

E, DD has the same strengthD(r) everywhere on thissurface.

Alan Murray – University of Edinburgh

EvaluateEvaluate ∫∫ ∫∫DD..dsds ∫∫ ∫∫ DD..dsds curved surface onlycurved surface only = = ∫∫ ∫∫ dsds = D= D∫∫ ∫∫ ds = D ds = D ×× area of curved surface area of curved surface = D = D × 2 π × 2 π rr LL So 2DSo 2Dπ π rr L = charge enclosedL = charge enclosed Charge enclosed?Charge enclosed? Charge/length Charge/length × × length L = length L = ρρ l l ×× LL

ρ l Coulombs/m

rL

22ππrr

DD

Alan Murray – University of Edinburgh

EvaluateEvaluate ∫∫ ∫∫DD..dsds ∫∫ ∫∫ DD..dsds = charge enclosed = charge enclosed 2π2πDrDr×× LL = = ρρ l l ×× LL D(r) = D(r) = ρρ ll

2π2πrr

DD(r) =(r) =ρρ l l âârr

2π2πrr

Alan Murray – University of Edinburgh

DiscussionDiscussion

||DD| is proportional to 1/r| is proportional to 1/r• Gets weaker with distanceGets weaker with distance• Intuitively correctIntuitively correct

DD points radially outwards ( points radially outwards (âr) ||DD| is proportional to | is proportional to ρρ ll

• More charge density = more fieldMore charge density = more field• Intuitively correctIntuitively correct

Alan Murray – University of Edinburgh

Other forms of charge distribution?Other forms of charge distribution?

Spherical charge Spherical charge distributiondistribution

ρρ αα rr-2-2, r, r-3-3, e , e –r–r … … Choose a spherical Choose a spherical

surface for integrationsurface for integration Then Then DD and and dsds will will

once again be parallel once again be parallel on the surfaceon the surface

Check it out!Check it out!

r

Alan Murray – University of Edinburgh

`

Other forms of charge distribution?Other forms of charge distribution?

Sheet of chargeSheet of charge Mirror symmetryMirror symmetry Choose a surface that Choose a surface that

is symmetric about is symmetric about the sheetthe sheet

Then Then DD and and dsds will will once again be parallel once again be parallel or perpendicular on or perpendicular on the surfacesthe surfaces

Check it out!Check it out!

`

Alan Murray – University of Edinburgh

Gauss’s LawGauss’s Law

. ρ= =∫∫ ∫∫∫Dds enclosed vQ dv

This is Maxwell’s first equation

. 0=∫∫BdsAs there is not such thing as an isolated “magnetic charge”, no Gaussian surface can ever contain a net “magnetic charge” – they come in pairs (North and South poles).

And we can have Maxwell’s second equation for free!